Let the point (-1, α, β) lie on the line of the shortest distance between the lines (x+2)/-3 = (y-2)

Q: Wait, why can't we just use the shortest distance formula $\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$?

That is a great question! That formula is perfect for finding the *length* of the shortest distance, but this problem asks for a point *on* the line of the shortest distance. To find a point, we need the actual equation of the line, which requires finding the specific points $P$ and $Q$ where the shortest distance line intersects the two given lines. The formula alone won't give you those coordinates.

Q: Why do we equate the direction ratios of $\vec{PQ}$ and the cross product vector $\vec{n}$?

Because $\vec{PQ}$ is the vector representing the shortest distance line itself! Since the shortest distance line is perpendicular to both given lines, it must be parallel to the vector $\vec{n} = \vec{b}_1 \times \vec{b}_2$. When two vectors are parallel, their direction ratios are proportional. Equating these ratios is the standard algebraic way to enforce that parallelism.

Q: Is it possible for the lines to intersect? How would that change the problem?

If the lines intersected, the shortest distance would be zero. In that case, the vector $\vec{PQ}$ would be the zero vector at the point of intersection. However, in JEE problems of this type, the lines are almost always skew. You can verify this by checking if the shortest distance is non-zero or by checking if the determinant of the vectors $(\vec{a}_2 - \vec{a}_1, \vec{b}_1, \vec{b}_2)$ is non-zero.

Question

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3} = \frac{y-2}{4} = \frac{z-5}{2}$ and $\frac{x+2}{-1} = \frac{y+6}{2} = \frac{z-1}{0}$. Then $(\alpha - \beta)^2$ is equal to___________

Sigma Percentile · Accepted Answer

### The Geometry of Connection: Navigating Skew Lines

Imagine you are standing in a vast, three-dimensional void. In front of you, there are two infinite lines, $L_1$ and $L_2$. They are not parallel, yet they never touch.

They are skew lines, dancing around each other in the infinite expanse of space. Your mission is to find the unique bridge that connects them—the line of shortest distance.

### Phase 1: The Setup

First, we must understand the anatomy of our lines. We are given:

$$L_1: \frac{x+2}{-3} = \frac{y-2}{4} = \frac{z-5}{2}$$

$$L_2: \frac{x+2}{-1} = \frac{y+6}{2} = \frac{z-1}{0}$$

By inspecting the denominators, we extract their direction vectors: $\vec{b}_1 = -3\hat{i} + 4\hat{j} + 2\hat{k}$ and $\vec{b}_2 = -1\hat{i} + 2\hat{j} + 0\hat{k}$. These vectors define the orientation of our lines in space.

### Phase 2: The Common Perpendicular

To find the shortest distance line, we need a direction that is perpendicular to both $\vec{b}_1$ and $\vec{b}_2$. We calculate the cross product $\vec{n} = \vec{b}_1 	imes \vec{b}_2$.

Expanding the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 4 & 2 \ -1 & 2 & 0 \end{vmatrix} = -4\hat{i} - 2\hat{j} - 2\hat{k}$$

To simplify, we scale this vector to $\vec{n} = (2, 1, 1)$. This vector is the direction of our bridge.

### Phase 3: The Parametric Dance

Now, we define the points $P$ and $Q$ where this bridge touches our lines. We express $P$ on $L_1$ and $Q$ on $L_2$ using parameters $\lambda$ and $\mu$:

$$P(-3\lambda-2, 4\lambda+2, 2\lambda+5)$$

$$Q(-\mu-2, 2\mu-6, 1)$$

The vector $\vec{PQ}$ connecting them is:

$$\vec{PQ} = (3\lambda - \mu)\hat{i} + (2\mu - 4\lambda - 8)\hat{j} + (-2\lambda - 4)\hat{k}$$

### Phase 4: The Parallel Condition

Since $\vec{PQ}$ lies along the shortest distance line, it must be parallel to our direction vector $\vec{n} = (2, 1, 1)$. This gives us the proportionality condition:

$$\frac{3\lambda-\mu}{2} = \frac{2\mu-4\lambda-8}{1} = \frac{-2\lambda-4}{1}$$

Solving this system of equations yields the values **$\lambda = -1$** and **$\mu = 1$**.

### Phase 5: The Final Destination

With $\lambda = -1$, we find the point $P$ to be $(1, -2, 3)$. The equation of the shortest distance line is:

$$\frac{x-1}{2} = \frac{y+2}{1} = \frac{z-3}{1}$$

The problem states that the point $(-1, \alpha, \beta)$ lies on this line. Substituting $x = -1$:

$$\frac{-1-1}{2} = -1$$

Equating this to the other components:

$$\frac{\alpha+2}{1} = -1 \Rightarrow \alpha = -3$$

$$\frac{\beta-3}{1} = -1 \Rightarrow \beta = 2$$

Finally, we calculate the result:

$$(\alpha - \beta)^2 = (-3 - 2)^2 = (-5)^2 = **25**$$

Enter Numerical Value:

Visualized Solution

Identifying the Given Lines

Direction of the Shortest Distance Line

Calculating the Cross Product

Parametric Coordinates of $P$ and $Q$

Defining Vector $P Q$

Applying the Parallel Condition

Solving for $λ$ and $μ$

Finding Coordinates of $P$

Equation of the Shortest Distance Line

Substituting the Target Point

Final Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Connection

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______.

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{5}$ , then the sum of all possible values of $λ$ is :

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{29}$ , then a value of $λ$ is :

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :